# Problems for the week

#### Problem 1.

Given any set of points within a unit square, show that there always exists three distinct points in such that the area of the triangle formed by the three points is less than or equal to . (Combinatorics)

#### Problem 2.

For every positive integer , prove that where is the sum of all the divisors of . (Number Theory)

#### Problem 3.

Does there exist non-linear polynomials and such that (Polynomials)

#### Problem 4.

If is a twice differentiable function and is continuous on an open interval in , then prove that (Calculus)

### Go to Week 2 to challenging problems: Click Here

• Shakshi raj

Thanks sir

• Sounak Das

Sir..can i submit solution??

• Yes, you can type them here.

• Chowman

For the first problem , pigeonhole principle solves the problem quite easily .
We divide the given square into smaller squares of sides of length each . By pigeonhole principle , we can say that when 9 points are divided amongst 4 smaller squares , one sqaure must contain points .
Let’s call the square M . In the sqaure M, we join any of the diagonals , and we have a right angle triangle whose maximum base is and maximum height is So we can conlude that the area has an upper bound of • Yes, you are correct but the result that you are using for the maximum area of a triangle inside a square has to be explicitly explained as to why some other triangles will not have area more than where is the side length of the square.

• Chowman

For the second problem
We start by plugging in values .Like for We notice that all of the fractions are in the form of .
Then we have the fraction occurring times.

Note that , we are basically counting the number of times 1/k occurs.

So the sum And thus proved .

Although the bound is probably even less than • Chowman

The latex equation got messed up in this one .

• I think it’s fine now. Thanks a lot for the discussion. I will comment on the posts tomorrow early morning. Till then solve the rest.

• You are absolutely correct. Well done.